q=mc(deltaT) final temp

Question

A hot lump of 45.345.3 g of aluminum at an initial temperature of 83.6 °C83.6 °C is placed in 50.0 mL H2OH2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)?0.903 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

heat lost by hot aluminum MUST equal heat gained by cooler H2O. The equation, as indicated in the title, is

q = mC∆T

Heat lost by aluminum = q = mC∆T = (45.3 g)(0.903 J/gº)(83.6º - Tf) where Tf = final temperature

Heat gained by H2O = q = mC∆T = (50.0 g)(4.184 J/gº)(Tf - 25.0º)

Setting these two equal and solving for Tf, we have...

(45.3 g)(0.903 J/gº)(83.6º - Tf) = (50.0 g)(4.184 J/gº)(Tf - 25.0º)

3420 - 40.9Tf = 209.2Tf - 5230

250.1Tf = 8650

Tf = 34.6ºC

(be sure to check the math)